Use the van\'t Hoff factors in the table below to calculate each colligative pro

Question

Use the van't Hoff factors in the table below to calculate each colligative property.

Table. Van't Hoff Factors at 0.05 m Concentration in Aqueous Solution

Part A

the melting point of a 0.100 m iron(III) chloride solution

Express your answer using three significant figures.

Part B

the osmotic pressure of a 0.080 M potassium sulfate solution at 298 K

Express your answer using two significant figures.

Part C

the boiling point of a 1.34 % by mass magnesium chloride solution

Express your answer using two decimal places.

Solute i Expected i Measured Nonelectrolyte 1 1 NaCl 2 1.9 MgSO4 2 1.3 MgCl2 3 2.7 K2SO4 3 2.6 FeCl3 4 3.4

Explanation / Answer

Solution :-

Part A

the melting point of a 0.100 m iron(III) chloride solution

FeCl3 have i = 3.4

So lets calculate the delta T

Delta T = Kf * m * i

             = 1.86 C per m * 0.100 m * 3.4

             = 0.6324 C

Freezing point = 0 – 0.6324 C = -0.6324 C

So the melting point of solution is -0.6324 C

Part B

the osmotic pressure of a 0.080 M potassium sulfate solution at 298 K

formula to calculate the osmotic pressure is

pi = m*r*T*I

   = 0.080 m * 0.08206 L atm per mol K * 298 K * 2.6

   = 5.09 atm

Part C

the boiling point of a 1.34 % by mass magnesium chloride solution

solution :-

1.34 % MgCl2 means 1.34 g in 100 g solution

Mass of water = 100 g – 1.34g =   98.66 g

So moles of MgCl2 = mass / molar mass

                               = 1.34 g / 95.211 g per mol

                               = 0.01407 mol

Molality = moles / kg solvent

                = 0.01407 mol / 0.09866 kg

                = 0.1426 m

Calculating the delta Tb

DeltaTb = Kb * m* i

               = 0.512 C per m * 0.1426 m * 2.7

               = 0.200 C

So the boiling point of solution = 100 C + 0.197 C = 100.2 C

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